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Old 10-13-2013, 09:59 PM
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yaser yaser is offline
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Default LR and PLA with scaled input space

A post at another forum:
While answering question 7, I mistakenly took my test points from [0,1]\times[0,1], which resulted in the number of iterations being approximately double of what it was when I corrected it to [-1,1]\times[-1,1].

I thought that this might simply be because doubling the interval provided more area, making the points more loosely scattered and allowing for a larger number of lines to satisfy the test cases. However, when I tested my hypothesis by taking intervals of [-2,2]\times[-2,2] and [-4,4]\times[-4,4], etc., I found that the number of iterations was least for [-2,2]\times[-2,2] and went up in either direction from there. The points which defined my line were also taken from the larger intervals.

Can anyone help explain why this might be the case?
If you scale \bf x, then the linear regression solution \bf w scales in the opposite direction (other things being equal) since it is trying to make {\bf w}^{\rm T} {\bf  x} match the same value (+1 or -1). Now if you take the LR solution \bf w and use it as initial condition for PLA, the impact of each PLA iteration scales up with \bf x since you are adding (y {\bf x}) to the weight vector at each iteration.

Put these together and you conclude that, as \bf x scales up and down, the impact of the LR solution vector on PLA goes down and up, respectively, and significantly so. On the large \bf x extreme, the LR solution \bf w behaves like the vector \bf 0 so you get the original PLA iterations. As \bf x gets smaller, \bf w kicks in as a good initial condition (with non-trivial size) and you gain some PLA iterations. As \bf x diminishes, PLA will take longer to correct the misclassified points that the LR \bf w didn't get simply because the PLA iteration becomes relatively smaller in the movement that it creates.
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